An object is projected from ground level at an angle of 60^\circ to the horizontal. If the… — JAMB UTME

An object is projected from ground level at an angle of $60^\circ$ to the horizontal. If the horizontal component of its initial velocity is $25\text{ m/s}$ , calculate the maximum height reached by the object in meters. (Take $g = 10\text{ m/s}^2$ )

Cevap: 93.75 m

Cevap

The maximum height reached by the object is

93.75\text{ m}

The maximum vertical height attained by a projectile depends on its vertical velocity component

u_y = u \sin \theta

. Resolving the initial velocity gives

u = 50\text{ m/s}

and

u_y = 25\sqrt{3}\text{ m/s}

. Substituting into

H = \frac{u_y^2}{2g}

yields

93.75\text{ m}

Adım Adım Çözüm

Find the magnitude of the initial velocity

u = 50\text{ m/s}

The horizontal velocity component remains constant throughout flight and is given by

u_x = u \cos \theta

Calculate the initial vertical velocity component

u_y = 25\sqrt{3}\text{ m/s}

Vertical component of velocity is calculated using

u_y = u \sin \theta

Calculate the maximum height

H = 93.75\text{ m}

At maximum height, vertical velocity is zero, giving

H = \frac{u_y^2}{2g}

Anahtar Kavram

Resolution of velocity components in projectile motion and calculation of maximum height

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